JEE Main 2023DifferentiationDifferentiationEasyQuestiond2xdy2{{{d^2}x} \over {d{y^2}}}dy2d2x equals:OptionsA−(d2ydx2)−1(dydx)−3 - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}−(dx2d2y)−1(dxdy)−3B(d2ydx2)(dydx)−2{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}(dx2d2y)(dxdy)−2C−(d2ydx2)(dydx)−3 - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}−(dx2d2y)(dxdy)−3D(d2ydx2)−1{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}(dx2d2y)−1Check AnswerHide SolutionSolutiond2xdy2=ddy(dxdy){{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)dy2d2x=dyd(dydx) =ddx(dxdy)dxdy = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}=dxd(dydx)dydx =ddx(1dy/dx)dxdy = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}=dxd(dy/dx1)dydx =−1(dydx)2.d2ydx2.1dydx = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}=−(dxdy)21.dx2d2y.dxdy1 =−1(dydx)3d2ydx2 = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}=−(dxdy)31dx2d2y